What is the integral of (e^x+1/(e^{2x)})^2 ?

The integral of (e^x+1/(e^{2x)})^2 is 1/2 e^{2x}-2/(e^x)-1/4 e^{-4x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^x+1/(e^{2x)})^2

\int (e^{x} + \frac{1}{e ^{2 x}})^{2} d x

\frac{1}{2} e^{2 x} - \frac{2}{e ^{x}} - \frac{1}{4} e^{- 4 x} + C

Solution steps

\int (e^{x} + \frac{1}{e ^{2 x}})^{2} d x

Expand

(e^{x} + \frac{1}{e ^{2 x}})^{2} : e^{2 x} + \frac{2}{e ^{x}} + \frac{1}{e ^{4 x}}

= \int e^{2 x} + \frac{2}{e ^{x}} + \frac{1}{e ^{4 x}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{2 x} d x + \int \frac{2}{e ^{x}} d x + \int \frac{1}{e ^{4 x}} d x

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

\int \frac{2}{e ^{x}} d x = - \frac{2}{e ^{x}}

\int \frac{1}{e ^{4 x}} d x = - \frac{1}{4} e^{- 4 x}

= \frac{1}{2} e^{2 x} - \frac{2}{e ^{x}} - \frac{1}{4} e^{- 4 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} - \frac{2}{e ^{x}} - \frac{1}{4} e^{- 4 x} + C

What is the integral of (e^x+1/(e^{2x)})^2 ?
The integral of (e^x+1/(e^{2x)})^2 is 1/2 e^{2x}-2/(e^x)-1/4 e^{-4x}+C